Rank, Nullity and Elimination
• Last Frame Test
• Last Frame Example
• Definition: Reduced Echelon System
• Definition: Rank and Nullity
• Theorem: Rank plus Nullity equals Variable Count
• Elimination
– Theorem: Elimination
– An Elimination Algorithm
– Example
Last Frame Test
A frame in a sequence without a signal equation 0 = 1 is called the Last Frame provided:
• Each nonzero equation has a lead variable.
• Equations 0 = 0 appear at the end.
• Variables within an equation appear in variable list order.
• Nonzero equations are listed with lead variables in variable list order.
Last Frame Example
Let’s assume variable list order x, y, z .

x + y

= 2,
z = 3,
0 = 0.
The lead variables are x, z and the free variable is y .
Definition 1 (Reduced Echelon System)
A linear system which passes the last frame test is called a reduced echelon system.
Definition 2 (Rank and Nullity)
Assume the last frame test has been passed. Then
• Rank = number of lead variables,
• Nullity = number of free variables (non-lead variables).
Determining the rank and nullity of a system
• Display a frame sequence whose first frame is that system, and whose last frame passes
the Last Frame Test. The last frame is a reduced echelon system.
• The rank of the system is the number of lead variables in the last frame.
• The nullity of the system is the number of variables minus the rank.
Theorem 1 (Rank and Nullity)
The following equation holds:
rank + nullity = number of variables.
Elimination
The elimination algorithm applies at each algebraic step one of the three toolkit rules swap,
multiply and combination.
• The objective of each algebraic step is to increase the number of lead variables. The
process stops when a signal equation (typically 0 = 1) is found. Otherwise, it stops
when no more lead variables can be found, and then the last system of equations is a
reduced echelon system. A detailed explanation of the process has been given in the
discussion of frame sequences.
• Reversibility of the algebraic steps means that no solutions are created nor destroyed
throughout the algebraic steps: the original system and all systems in the intermediate
steps have exactly the same solutions.
• The final reduced echelon system has either a unique solution or infinitely many solutions. In both cases we report the general solution. In the infinitely many solution
case, the last frame algorithm is used to write out a general solution.
Theorem 2 (Elimination)
Every linear system has either no solution or else it has exactly the same solutions
as an equivalent reduced echelon system, obtained by repeated application of the
toolkit rules swap, multiply and combination.
An Elimination Algorithm
An equation is said to be processed if it has a lead variable. Otherwise, the equation is said
to be unprocessed.
1. If an equation “0 = 0” appears, then move it to the end. If a signal equation
“0 = c” appears (c 6= 0 required), then the system is inconsistent. In this
case, the algorithm halts and we report no solution.
2. Identify the first symbol xr , in variable list order x1 , . . . , xn , which appears
in some unprocessed equation. Apply the multiply rule to insure xr has
leading coefficient one. Apply the combination rule to eliminate variable xr
from all other equations. Then xr is a lead variable: the number of lead
variables has been increased by one.
3. Apply the swap rule repeatedly to move this equation past all processed
equations, but before the unprocessed equations. Mark the equation as
processed, e.g., replace xr by boxed symbol xr .
4. Repeat steps 1–3, until all equations have been processed once. Then lead
variables xi1 , . . . , xim have been defined and the last system is a reduced
echelon system.
1 Example (Elimination) Solve the system.
w + 2x − y + z = 1,
w + 3x − y + 2z = 0,
x
+ z = −1.
Solution
The answer using the natural variable list order w , x, y , z is the standard general solution
w
x
y
z
=
=
=
=
3 + t1 + t2,
−1 − t2,
t1,
t2,
Details appear in the next few slides.
−∞ < t1, t2 < ∞.
Elimination Details
We will apply the three rules swap, multiply and combination for equivalent equations to
obtain a frame which passes the Last Frame Test. Then we apply the Last Frame Algorithm
to obtain the general solution of the system.
Let’s mark processed equations with a box enclosing the lead variable (w is marked w ).
w + 2x − y + z = 1
w + 3x − y + 2z = 0
x
+ z = −1
1
w + 2x − y + z = 1
0 + x + 0 + z = −1
x
+ z = −1
2
w + 2x − y + z = 1
x
+ z = −1
0 = 0
3
w + 0 − y − z = 3
x
+ z = −1
0 = 0
4
Documentation of the steps
1 Original system. Identify the variable order as w, x, y , z .
2 Choose w as a lead variable.
combo(1,2,-1).
Eliminate w from equation 2 by using
3 The w-equation is processed. Let x be the next lead variable. Eliminate x from
equation 3 using combo(2,3,-1).
4 Eliminate x from equation 1 using combo(2,1,-2). Mark the x-equation as processed. Last Frame Test passed.
The four frames make the frame sequence which takes the original system into the last
frame. Basic exposition rules apply:
1. Variables in an equation appear in variable list order.
2. Equations inherit variable list order from the lead variables.
Last Frame Algorithm
The last frame of the sequence, which must pass the Last Frame Test, is used to write out
the general solution, as follows.
w
x
y
z
= 3 + y + z
= −1 − z
= t1
= t2
Solve for the lead variables w , x. Assign invented symbols t1 , t2 to the free
variables y , z .
w
x
y
z
= 3 + t1 + t2
= −1 − t2
= t1
= t2
Back-substitute free variables into the
lead variable equations to display the
standard general solution.
Requirement: variables must be listed
in variable list order [w, x, y, z ].